248
7 Torsion of Prismatic Bars
2Gθ =
1
A 2
(a 2 q 2 − a 12 q 1 )
=
1
2000
(2409q 2 − 366q 1 )
Therefore,
2Gθ = 1.20q 2 − 0.18q 1
(ii)
Equating (i) and (ii), we get
2.18q 1 − 0.54q 2 = 1.20q 2 − 0.18q 1
or
2.36q 1 − 1.74q 2 = 0
or
q 2 = 1.36q 1
The torque due to shear flows should be equal to the applied torque
Hence, from Eq. (7.28),
M t = 2q 1 A 1 + 2q 2 A 2
10,000 × 100 = 2q 1 × 680 + 2q 2 × 2000
= 1360q 1 + 4000q 2
Substituting for q 2 , we get
10,000 × 100 = 1360q 1 + 4000 × 1.36q 1
Therefore,
q 1 = 147N and q 2 = 200N
Example 7.3 A thin-walled steel section shown in Fig. 7.13 is subjected to a twisting
moment T. Calculate the shear stresses in the walls and the angle of twist per unit
length of the box.
Solution Let A 1 and A 2 be the areas of the cells (1) and (2) respectively.
∴ A 1 =
πa
2
2
7 Torsion of Prismatic Bars
2Gθ =
1
A 2
(a 2 q 2 − a 12 q 1 )
=
1
2000
(2409q 2 − 366q 1 )
Therefore,
2Gθ = 1.20q 2 − 0.18q 1
(ii)
Equating (i) and (ii), we get
2.18q 1 − 0.54q 2 = 1.20q 2 − 0.18q 1
or
2.36q 1 − 1.74q 2 = 0
or
q 2 = 1.36q 1
The torque due to shear flows should be equal to the applied torque
Hence, from Eq. (7.28),
M t = 2q 1 A 1 + 2q 2 A 2
10,000 × 100 = 2q 1 × 680 + 2q 2 × 2000
= 1360q 1 + 4000q 2
Substituting for q 2 , we get
10,000 × 100 = 1360q 1 + 4000 × 1.36q 1
Therefore,
q 1 = 147N and q 2 = 200N
Example 7.3 A thin-walled steel section shown in Fig. 7.13 is subjected to a twisting
moment T. Calculate the shear stresses in the walls and the angle of twist per unit
length of the box.
Solution Let A 1 and A 2 be the areas of the cells (1) and (2) respectively.
∴ A 1 =
πa
2
2